求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2006)(b+2006)
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求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2006)(b+2006)
你这个题目少一个条件
再问: 已知|ab-2|与(b-1)2次方互为相反数
再答: |ab-2|与(b-1)2次方互为相反数 ∴ab-2=0 b-1=0 ∴a=2 b=1 ∴1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2006)(b+2006) =1/1×2+1/2×3+1/3×4+……+1/2007×208 =1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/2007-1/2008 =1-1/2008 =2007/2008
再问: 已知|ab-2|与(b-1)2次方互为相反数
再答: |ab-2|与(b-1)2次方互为相反数 ∴ab-2=0 b-1=0 ∴a=2 b=1 ∴1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2006)(b+2006) =1/1×2+1/2×3+1/3×4+……+1/2007×208 =1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/2007-1/2008 =1-1/2008 =2007/2008
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